Question

If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the following equation. h equals 128 t minus 32 t squared a. Find the maximum height attained by the object. b. Find the number of seconds it takes the object to hit the ground.

Answer 1

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

$h = 128t-32t^2$....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

$\dfrac{dh}{dt}=0$

From equation (I)

$\dfrac{dh}{dt}=128-64t$

$128-64t=0$

$t=\dfrac{128}{64}$

$t=2\ sec$

Now, for maximum height

Put the value of t in equation (I)

$h =128\times2-32\times4$

$h=128\ ft$

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

$128t-32t^2=0$

$t(128-32t)=0$

$128=32t$

$t=4\ sec$

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.